The larger of the two areas (in sq. units) into which the circle x 2 + y 2 = 16 a 2 is divided by the parabola y 2 = 6 a x , is

Step-by-step solution

For point of intersection, x 2 + 6 a x = 16 a 2 ⇒ x = 2 a Area = π ⋅ ( 4 a ) 2 − 2 ( ∫ 0 2 a 6 a x d x + ∫ 2 a 4 a 16 a 2 − x 2 d x ) = 16 π a 2 − 2 ( 2 6 a   x ⋅ 3 2 3 ) 0 2 a + ( x 2 16 a 2 − x 2 + 16 a 2 2 sin − 1 x 4 a ) 2 a 4 a = 16 π a 2 − 2 2 6 a ⋅ 2 2   a a 3 + ( 8 a 2 ⋅ π 2 − ( a ⋅ 2 3 a + 8 a 2 ⋅ π 6 ) ) = 16 π a 2 − 2 8 3 a 2 3 + 8 π a 2 3 − 2 3 a 2 =